Posted 15 Mar 2013 · Report post You have 50 coins that all look exactly the same. One of them is fake and weighs a few grams more than the others. Given a balance scale (not a triple beam scale), what is the fewest number of weightings you could use to find the fake coin? Explain. 0 Share this post Link to post Share on other sites

0 Posted 15 Mar 2013 (edited) · Report post You need 3 or 4 weighings Divide into 3 heaps - 17, 17, and 16 and weigh the first two. You now know which heap has the fake coin. Repeat process with the selected heap - 6,6,5 or 5,5, 6 as the case may be, and weigh the first two. Repeat again 2,2,2 or 2,2,1. If the single coin is fake, no further weighing is required. Else You have to do a 4th weighing if the fake coin is in a group of 2. Edited 15 Mar 2013 by bonanova Add spoiler 0 Share this post Link to post Share on other sites

0 Posted 15 Mar 2013 · Report post You have 50 coins that all look exactly the same. One of them is fake and weighs a few grams more than the others. Given a balance scale (not a triple beam scale), what is the fewest number of weightings you could use to find the fake coin? Explain. You have 50 possibilities Each weighing distinguishes among 3 cases. 3x3x3 = 27, so you'd need a fourth weighing for this case. 0 Share this post Link to post Share on other sites

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You have 50 coins that all look exactly the same. One of them is fake and weighs a few grams more than the others. Given a balance scale (not a triple beam scale), what is the fewest number of weightings you could use to find the fake coin? Explain.

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